I'm new to the theory of affine connections and have yet to become comfortable working with. I wanted to confirm to myself that the curvature of an affine connection is skew-symmetric which boils down to showing that $$\nabla_{[X,Y]}Z = -\nabla_{[Y,X]}Z$$
where $X,Y,Z$ are assumed to be smooth vector fields on a manifold $M$ and $[X,Y]$ denotes the Lie bracket.
Can we show this by using the $C^\infty$-linearity of $\nabla$ in the following way?
$$\nabla_{[X,Y]}Z = \nabla_{-[Y,X]}Z = \nabla_{-\operatorname{id}[Y,X]}Z = -\operatorname{id}\nabla_{[Y,X]}Z = -\nabla_{[Y,X]}Z $$
I.e. would this attempt be valid? I hope this does not come across as low effort, I just wanted to clarify wether this is the correct way to handle it.
Thank you very much!
Forget connections for one second: $[X,Y]=-[Y,X]$ from the definition of Lie bracket. Then $$\nabla_{[X,Y]}Z = \nabla_{-[Y,X]}Z = - \nabla_{[Y,X]}Z.$$Quoting $C^\infty$-linearity to pull out ${\rm id}$ is meaningless as this is not a real-valued function.